Chapter 5.5 Limits

In this section, we learn about what it means to take the limit of a function. We will use sequences to make sense of this idea.

Definition and Computation of Limits

Take \(f\) to be the function that is given by \[f(x) = \tfrac{x}{x}.\] The function \(f\) looks like this:

It is defined for all real numbers except zero. So, let’s view \(f\) as the restriction of the constant function \(1(x)=1\) on the domain \((-\infty, 0)\cup(0, \infty)\) so that \(1\) is an extension of \(f\) that is defined on \(\mathbb R\).

Goal: Make precise the intuition that \(f\) has a “hole” at the point \((0,1)\) that can be “filled in.”

Take \(f\) to be a real valued function that is defined on a subset of \(\mathbb R\) and take \(x_0\) to be a real number that may or may not be in the domain of \(f\).

The following statement about \(f\) makes precise the idea that \(f(x)\) is as close as we like to the real number \(L\) as long as \(x\) is in \(\mathcal D(f)\smallsetminus\{x_0\}\) and close enough to \(x_0\):

Limit of a function

The real number \(L\) is the limit of \(f(x)\) as \(x\) tends to \(x_0\), if for any sequence \((x_n)\) in \(\mathcal D(f)\smallsetminus\{x_0\}\) that converges to \(x_0\), \((f(x_n))\) converges to \(L\).

Express this symbolically as: \(\displaystyle \lim_{x\to x_0} f(x) = L\).

Use this picture to make sense of the idea.

Understand how to take a limit of a function with this example.

Example 1

Take \(f\) and \(g\) to be the functions that are given by \[f(x) = x+2\quad {\rm and}\quad g(x) = \begin{cases}\frac{x^2+x-2}{x-1} &\text{if }x\ne 1\\7&\text{if }x = 1\end{cases}\] and determine \[\lim_{x\to 1}f(x) \quad {\rm and} \quad \lim_{x\to 1}g(x).\] Does the second limit depend on the value \(g(1)\)?

1. Take \((x_n)\) to be a sequence in \(\mathcal D(f)\smallsetminus\{1\}=(-\infty,1)\cup(1,\infty)\) that converges to \(1\). Then \(f(x_n)=x_n+2\) and \(g(x_n)=\frac{(x_n)^2+x_n-2}{x_n-1}\), Use the theorems from 5.3 to get \[\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}\left(x_n+2\right)=\lim_{n\to\infty}x_n+\lim_{n\to\infty}2=1+2=3.\] Since this worked for any sequence in \(\mathcal D(f)\smallsetminus\{1\}\), the conclusion is that \(\displaystyle\lim_{x\to 1}f(x)=3\).

2. Take \((x_n)\) to be a sequence in in \(\mathcal D(g)\smallsetminus\{1\}=(-\infty,1)\cup(1,\infty)\) that converges to \(1\), For \(g\), the numerator tends to \(0\) and the denominator tends to \(0\). So the limit laws cannot be used. Rewrite \(g(x_n)\) like this: \[\frac{(x_n)^2+x_n-2}{x_n-1}=\frac{(x_n-1)(x_n+2)}{x_n-1}=x_n+2,\] which is valid for all \(n\) since \(x_n\not=1.\) Therefore, \[\lim_{n\to\infty}g(x_n)=\lim_{n\to\infty}\left(\frac{(x_n)^2+x_n-2}{x_n-1}\right)=\lim_{n\to\infty}\left(x_n+2\right)=\lim_{n\to\infty}(x_n)+\lim_{n\to\infty}2=1+2=3.\] Since this worked for any sequence in \(D(g)\smallsetminus\{1\}\), the conclusion is that \(\displaystyle\lim_{x\to 1}g(x)=3\)

The second limit did not depend on the value of \(g(1)\).

Verification of the equality \[\lim_{x\to x_0}f(x) = L\] requires the verification of a statement about every sequence in the domain of \(f\) that is convergent to \(x_0\).

To show that \[\lim_{x\to x_0}f(x) \ne L\] requires demonstrating the existence of just a single sequence.

Namely, \(L\) is not the given limit if and only if there is a sequence \((x_n)\) in \(\mathcal D(f)\smallsetminus\{x_0\}\) that converges to \(x_0\), but \((f(x_n))\) does not converge to \(L\).

Practice using this definition with this example.

Example 2

Show that the following limits do not exist:

1. \(\displaystyle \lim_{x\to 1} \tfrac{|x-1|}{x-1}\);

2. \(\displaystyle \lim_{x\to 0} \sin\!\big(\tfrac{1}{x}\big)\).

A result from Chapter 5.4 states that a convergent sequence has a unique limit. Therefore, a limit does not exist if there are two different sequences \(x_n\) and \(y_n\) that converge to \(x_0\) and for which \(f(x_0)\) and \(f(y_n)\) converge, but to two different values.

1. Take \(x_n=1+\frac{1}{n}\). This sequence converges to \(1\). Note that \[ \begin{align*} |x-1|&=\begin{cases} x-1&\text{ if } x\geq 1\\ -(x-1)&\text{ if }x<1 \end{cases} \end{align*} \] implies that \[|x_n-1|=x_n-1.\] Therefore, since \(x_n\not=1\), we have that \[\frac{|x_n-1|}{x_n-1}=\frac{x_n-1}{x_n-1}=1\] and so \[\lim_{n\to\infty}\frac{|x_n-1|}{x_n-1}=\lim_{n\to\infty}=1=1.\] Now, take \(y_n=1-\frac{1}{n}\). This sequence converges to \(1\). Note that \[\begin{align*} |x-1|&=\begin{cases} x-1&\text{ if } x\geq 1\\ -(x-1)&\text{ if }x<1 \end{cases} \end{align*} \] implies that \[|y_n-1|=-(y_n-1).\] Therefore, since \(y_n\not=1\), we have that \[\frac{|y_n-1|}{y_n-1}=\frac{-(y_n-1)}{y_n-1}=-\] and so \[\lim_{n\to\infty}\frac{|y_n-1|}{y_n-1}=\lim_{n\to\infty}=-1=-1.\] Thus the limit does not exist at \(x=1\) for \(\tfrac{|x-1|}{x-1}.\)

2. Notice that if \((x_n)\) is a sequence that converges to \(0\), then \(\left(\frac{1}{x_n}\right)\) is sequence that can diverge to infinity or negative infinity. Take \((x_n)\) to be the sequence given by \(x_n=\frac{1}{n\pi}\). This is a null sequence and \[ \sin\left(\frac{1}{x_n}\right)=\sin\left(\frac{1}{\frac{1}{n\pi}}\right)=\sin(n\pi)=0 \]

for all \(n\). Thus \[\displaystyle\lim_{n\to\infty}\sin(x_n)=0.\]

Now take \((y_n)\) to be the sequence given by \(y_n=\frac{1}{\tfrac{\pi}{2}+2\pi n}\). This is also a null sequence and \[ \sin\left(\frac{1}{y_n}\right)=\sin\left(\frac{1}{\frac{1}{\tfrac{\pi}{2}+2\pi n}}\right)=\sin\left(\tfrac{\pi}{2}+2\pi n\right)=1 \]

for all \(n\). Thus \[\displaystyle\lim_{n\to\infty}\sin(y_n)=1.\]

We conclude that the limit does not exist at \(x=0\).

It is helpful to establish some initial results on limits of functions.

Limits of Pow Function and Constant Function

1. For any rational number \(r\), and any \(x_0\) at which \({\rm pow}_r\) is defined, \[\lim_{x\to x_0}{\rm pow}_r(x) = {\rm pow}_r(x_0), \quad \text{that is,} \quad \lim_{x\to x_0} x^r = x_0^r.\]

2. \(\displaystyle\lim_{x\to x_0}c=c\).

Understand this theorem with these properties.

Example 3

We have

1. \(\displaystyle \lim_{x\to 3}x=3\)

2. \(\displaystyle \lim_{x\to -1}x^2=(-1)^2=1\)

3. \(\displaystyle \lim_{x\to 0}x^{3/2}=(0)^{3/2}=0\)

In order to simplify statements about limits, we introduce the idea of a limit point.

Limit Point

For any subset \(X\) of \(\mathbb R\), \(x_0\) is a limit point of \(X\) if there is a sequence \((x_n)\) in \(X\smallsetminus\{x_0\}\) that converges to \(x_0\).

If \(x_0\) is in \(X\) but not a limit point of \(X\), then \(x_0\) is an isolated point of \(X\)

Understand the idea of a limit point with the figure below.

The limit laws for sequences imply these limit laws for real valued functions on \(\mathbb R\).

Limit Laws

For any functions \(f\) and \(g\) whose common domain \(D\) is nonempty, and for any limit point \(x_0\) of \(D\), if \[\lim_{x\to x_0} f(x) = L\quad {\rm and}\quad \lim_{x\to x_0} g(x) = M,\] then

1. (limit law for sums) \(\displaystyle\lim_{x\to x_0} (f+g)(x) = L+M\);

2. (limit law for products) \(\displaystyle\lim_{x\to x_0} (fg)(x) = LM\).

3. (limit law for quotients) If in addition \(M\) is not equal to \(0\), then \(\displaystyle\lim_{x\to x_0} \left(\tfrac{f}{g}\right)(x) = \tfrac{L}{M}\).

Use the limit laws to complete the example below.

Example 4

Take \(a\), \(b\), \(c\), and \(d\) to be functions that are defined in an open interval that contains \(9\) and \[\lim_{x\to 9} a(x) = 2,\quad \lim_{x\to 9} b(x) = 5,\quad \lim_{x\to 9} c(x) = -4,\quad \text{and} \quad \lim_{x\to 9} d(x) = 7.\] Use the limit laws to determine the following limits:

1. \(\displaystyle \lim_{x\to 9} \left(6a(x) - 3b(x)\right)\);

2. \(\displaystyle \lim_{x\to 9} \left(\tfrac{a(x)b(x)+2c(x)}{d(x)+1}\right)\).

1. Use the limit laws to get \[\begin{align*}\displaystyle \lim_{x\to 9} \left(6a(x) - 3b(x)\right)&=\lim_{x\to 9}6a(x)+\lim_{x\to 9}-3b(x)&&\text{limit law for sum}\\ &=\lim_{x\to 9}6\cdot\lim_{x\to 9}a(x)+\lim_{x\to 9}-3\cdot\lim_{x\to 9}b(x)&&\text{limit law for products}\\ &=6\cdot 2-3\cdot 5\\ &=12-15\\ &=-3. \end{align*}\]

2. The limit of the denominator is non-zero. Use the limit laws to get \[\begin{align*}\displaystyle \lim_{x\to 9} \left(\tfrac{a(x)b(x)+2c(x)}{d(x)+1}\right)&= \tfrac{\displaystyle\lim_{x\to 9}\Biggr( a(x)b(x)+2c(x)\Biggr)}{\displaystyle\lim_{x\to 9}(d(x)+1)} &&\text{limit law for quotients}\\ &= \tfrac{\displaystyle\lim_{x\to 9}\left( a(x)b(x)\right)+\displaystyle\lim_{x\to 9}(2c(x))}{\displaystyle\lim_{x\to 9}d(x)+\displaystyle\lim_{x\to 9}1} &&\text{limit law for sums}\\ &= \tfrac{\displaystyle\lim_{x\to 9} a(x)\cdot \displaystyle\lim_{x\to 9}b(x)+\displaystyle\lim_{x\to 9}2\cdot \displaystyle\lim_{x\to 9}c(x)}{\displaystyle\lim_{x\to 9}d(x)+\displaystyle\lim_{x\to 9}1} &&\text{limit law for products}\\ &=\frac{2\cdot 5+2\cdot(-4)}{7+1}\\ &=\frac{2}{8}\\ &=\frac{1}{4} \end{align*}.\]

Use the limit laws to determine the following limits. Pay attention to the difference in each of the parts in regards to when we use the limit laws and when we rewrite the expression.

Example 5

Determine the following limits and carefully justify your reasoning:

1. \({\displaystyle \lim_{x\to 3}\tfrac{x^2 +x+4}{2x-1}}\)

2. \({\displaystyle \lim_{x\to 2}\tfrac{(x-2)(2x-1)}{x^2-4}}\)

3. \(\displaystyle \lim_{h\to 0}\tfrac{(2+h)^2-4}{h}\)

1. Use the limit laws to calculate the limit. We have \[ \begin{align*} \displaystyle \lim_{x\to 3}\tfrac{x^2 +x+4}{2x-1}&=\frac{\displaystyle \lim_{x\to 3}(x^2+x+4)}{\displaystyle \lim_{x\to 3}(2x-1)}&&\text{limit law for quotients}\\ &=\frac{\displaystyle \lim_{x\to 3}x^2+\displaystyle \lim_{x\to 3}x+\displaystyle \lim_{x\to 3}4}{\displaystyle \lim_{x\to 3}2\cdot \displaystyle \lim_{x\to 3}x+\displaystyle \lim_{x\to 3}-1}&&\text{limit law for sums and products}\\ &=\frac{3^2+3+4}{2\cdot 3-1}\\ &=\frac{9+3+4}{5}\\ &=\frac{16}{5} \end{align*}\]

2. The quotient limit law cannot be used because the limit of the denominator is zero. However, the limit of the numerator is also zero, so rewrite the expression to identify the common zero. Notice that for \(x\not=2,\) \[\frac{(x-2)(2x-1)}{x^2-4}=\frac{(x-2)(2x-1)}{(x-2)(x+2)}=\frac{2x-1}{x+2}.\] Because the definition of the limit restricts to sequences for which \(x_n\not=2\), use this rewritten expression to get that \[\begin{align*} \lim_{x\to 2}\frac{(x-2)(2x-1)}{x^2-4}&=\lim_{x\to 2}\frac{2x-1}{x+2}\\ &=\frac{\displaystyle\lim_{x\to 2}(2x-1)}{\displaystyle\lim_{x\to 2}(x+2)}&&\text{ limit law for quotients}\\ &=\frac{\displaystyle\lim_{x\to 2}2\cdot \displaystyle\lim_{x\to 2}x+\displaystyle\lim_{x\to 2}-1}{\displaystyle\lim_{x\to 2}x+\displaystyle\lim_{x\to 2}2}&&\text{ limit law for sums and products}\\ &=\frac{2\cdot 2-1}{2+2}\\ &=\frac{3}{4}. \end{align*} \]

3. The quotient limit law cannot be applied here either because the limit of the denominator is zero. However, the limit of the numerator is also zero, so rewrite the expression to identify the common zero. Notice that for \(h\not=0,\) \[\frac{(2+h)^2-4}{h}=\frac{4+4h+h^2-4}{h}=\frac{4h+h^2}{h}=4+h.\] This implies then that \[\displaystyle \lim_{h\to 0}\frac{(2+h)^2-4}{h}=\lim_{h\to 0}(4+h)=4.\]

We have already shown that if \((a_n)\) is a null sequence with nonzero terms, then \((\sin(a_n))\) is a null sequence, \((\cos(a_n))\) converges to 1, and \[\lim_{n\to \infty} \tfrac{\sin(a_n)}{a_n} = 1\quad {\rm and}\quad \lim_{n\to \infty} \tfrac{1-\cos(a_n)}{a_n} = 0.\]

Since these limits are valid for any null sequence with nonzero terms, we have the following.

Special Limits of Sine and Cosine

\[\lim_{x\to 0} \sin(x) = 0, \quad \lim_{x\to 0} \cos(x) = 1,\quad \lim_{x\to 0} \tfrac{\sin(x)}{x} = 1,\quad {\rm and}\quad \lim_{x\to 0}\tfrac{1-\cos(x)}{x} = 0.\]

All other limits involving trigonometric functions can be determined from these limits.

In particular, if \(x_n\) is a sequence that is convergent to \(x_0\), then \(x_n-x_0\) is a null sequence. By using the angle addition laws, we have that

\[ \begin{align*} \lim_{n\to\infty}\sin(x_n)&=\lim_{n\to\infty}\sin(x_n-x_0+x_0)\\ &=\lim_{n\to\infty}\biggr(\sin(x_n-x_0)\cos(x_0)+\sin(x_0)\cos(x_n-x_0)\biggr)\\ &=\lim_{n\to\infty}\biggr(\sin(x_n-x_0)\cos(x_0)\biggr)+\lim_{n\to\infty}\bigg(\sin(x_0)\cos(x_n-x_0)\biggr)\\ &=0\cdot \cos(x_0)+\sin(x_0)\cdot 1\\ &=\sin(x_0) \end{align*} \]

and

\[ \begin{align*} \lim_{n\to\infty}\cos(x_n)&=\lim_{n\to\infty}\cos(x_n-x_0+x_0)\\ &=\lim_{n\to\infty}\biggr(\cos(x_n-x_0)\cos(x_0)+\sin(x_0)\sin(x_n-x_0)\biggr)\\ &=\lim_{n\to\infty}\biggr(\cos(x_n-x_0)\cos(x_0)\biggr)+\lim_{n\to\infty}\bigg(\sin(x_0)\sin(x_n-x_0)\biggr)\\ &=1\cdot \cos(x_0)+\sin(x_0)\cdot 0\\ &=\cos(x_0). \end{align*} \]

Since \(\tan(x)=\frac{\sin(x)}{\cos(x)},\) we can use our limit laws for quotients to calculate the limits of tangent. We summarize everything in the theorem below.

Sine, Cosine, and Tangent Limits

For any real number \(x_0\),

1. \(\displaystyle \lim_{x\to x_0}\sin(x)=\sin(x_0)\)

2. \(\displaystyle \lim_{x\to x_0}\cos(x)=\cos(x_0)\)

3. \(\displaystyle \lim_{x\to x_0}\tan(x)=\tan(x_0)\), so long as \(x_0\) is not an odd multiple of \(\tfrac{\pi}{2}.\)

Use the above theorem to compute the following limits.

Example 6

Determine the following limits and carefully justify your reasoning:

1. \({\displaystyle \lim_{x\to 0}\tfrac{\sin(5x)}{2x}}\);

2. \({\displaystyle \lim_{x\to 1}\tfrac{\tan(7(x-1))}{3(x-1)}}\)

A result from Chapter 5.4 states that for any null sequence \(x_n\), \(\displaystyle\lim_{n\to \infty}\frac{\sin(x_n)}{x_n}=1\). Therefore, \(\displaystyle\lim_{x\to 0}\frac{\sin(x)}{x}=1.\)

The quotient limit law cannot be used in both parts because the limit of the denominator is zero. However, the limit of the numerator is also zero.

1. Rewrite the expression in order to use the limit laws. We have that \[ \begin{align*} \displaystyle \lim_{x\to 0}\frac{\sin(5x)}{2x}&=\lim_{x\to 0}\frac{\sin(5x)}{2x}\cdot\frac{5}{5}\\ &=\lim_{x\to 0}\frac{5}{2}\cdot\frac{\sin(5x)}{5x}\\ &=\lim_{x\to 0}\frac{5}{2}\cdot\lim_{x\to 0}\frac{\sin(5x)}{5x}&&\text{ limit law for products}\\ &=\frac{5}{2}\cdot 1\\ &=\frac{5}{2}. \end{align*} \]

2. Notice that \(\tan(7(x-1))=\frac{\sin(7(x-1))}{\cos(7(x-1))}.\) Use this and the limit laws to calculate the limit. We have that \[\begin{align*} \displaystyle \lim_{x\to1}\frac{\tan(7(x-1))}{3(x-1)}&=\lim_{x\to 0}\frac{\sin(7(x-1))}{\cos(7(x-1))\cdot 3(x-1)}\\ &=\lim_{x\to 1}\frac{\sin(7(x-1))}{\cos(7(x-1))\cdot 3(x-1)}\cdot \frac{7}{7}\\ &=\lim_{x\to 1}\frac{7}{3\cos(7(x-1))}\cdot \frac{\sin(7(x-1))}{7(x-1)}\\ &=\lim_{x\to 1}\frac{7}{3\cos(7(x-1))}\cdot\lim_{x\to 1} \frac{\sin(7(x-1))}{7(x-1)}&&\text{ limit law for products}\\ &=\frac{7}{3\cos(7(1-1))}\cdot 1\\ &=\frac{7}{3}\cdot 1\\ &=\frac{7}{3} \end{align*} \]

The squeeze theorem for sequences implies an analogous theorem for functions that has the same name.

Squeeze Theorem

Squeeze Theorem: For any functions \(f\), \(g\), and \(h\) whose common domain \(D\) is nonempty, and for any limit point \(x_0\) of \(D\), if for every \(x\) in \(D\), \[f(x) \leq g(x)\leq h(x),\] and if there is a real number \(L\) so that \[\lim_{x\to x_0}f(x) = \lim_{x\to x_0} h(x) = L,\quad {\rm then} \quad \lim_{x\to x_0}g(x)=L.\]

Practice with the next example.

Example 7

Use the squeeze theorem to determine the following limit: \[\lim_{x\to 1}\; (x-1)^2\sin\big(\tfrac{1}{x-1}\big).\]

The limit laws for products does not apply because \(\sin\big(\tfrac{1}{x-1}\big)\) has no limit at \(x=1.\) Instead, use the Squeeze Theorem. Take \(g(x)=(x-1)^2\sin\big(\tfrac{1}{x-1}\big),\) The goal is to find two functions \(f\) and \(h\) for which \(f(x)\leq g(x)\leq h(x)\) for each \(x\) in the common domain \(D\) and to show \(\displaystyle \lim_{x\to 1}f(x)=L=\lim_{x\to 1}h(x)\).

1. Since \(-1\leq \sin(X)\leq 1\) for any real number \(X\), \(-1\leq\sin\big(\tfrac{1}{x-1})\leq 1\) for any real number \(x\not=1\). Since \((x-1)^2\) is always positive for \(x\not=1\), what follows is true : \[\begin{align*} -1&\leq\sin\big(\tfrac{1}{x-1})\leq 1\\ -(x-1)^2&\leq(x-1)^2\sin\big(\tfrac{1}{x-1})\leq (x-1)^2. \end{align*} \] Take \(f(x)=-(x-1)^2\) and \(h(x)=(x-1)^2\). The above calculations demonstrate that \(f(x)\leq g(x)\leq h(x)\) for all \(x\not=1\).

2. Use the limit laws to get \(\displaystyle\lim_{x\to 1}f(x)=\lim_{x\to 1}-(x-1)^2=0\) and \(\displaystyle\lim_{x\to 1}h(x)=\lim_{x\to 1}(x-1)^2=0\).

By the squeeze theorem, \[\lim_{x\to 1} (x-1)^2\sin\big(\tfrac{1}{x-1}\big)=0.\]

In the next example, we look at a limit of a composite function.

Example 8

Take \(f\) and \(g\) to be the functions that are given by \[f(x) = \begin{cases}3x+1&\text{if } x\ne 1\\-2&\text{if } x = 1\end{cases} \quad \text{and}\quad g(x) = 1.\] Determine the limits \[\lim_{x\to 0} g(x), \quad \lim_{x\to 1} f(x),\quad {\rm and}\quad \lim_{x\to 0}(f\circ g)(x).\] Does this answer surprise you at all?

Note that for all \(x,\) \[(f\circ g)(x)=f(g(x))=f(1)=-2,\]

Therefore, \[\lim_{x\to 0} g(x)=1,\] \[\lim_{x\to 1} f(x)=\lim_{x\to 1}(3x+1)=4,\] and \[\lim_{x\to 0} (f\circ g)(x)=-2.\]

This means that \[\lim_{x\to 0} (f\circ g)(x)\not=f(\lim_{x\to 0}g(x)).\]

This is not surprising because the limit as \(x\) approaches \(1\) of \(f\) does not depend on the value of \(f(1).\) On the other hand, \(\lim\limits_{x\to 0}(f\circ g)(x)\) only depends on \(f(1)\) and not \((f\circ g)(0).\)

In a future section, we will learn about a limit law for composite functions. For now, use the next example to understand how to use factorization of polynomials to evaluate limits.

Example 9

Calculating limits can often require some subtle manipulations. Here is an example of how one can use clever factorizations of polynomials to determine limits:

1. \(\lim\limits_{h\to0}\tfrac{\sqrt{4+h}-2}{h}\)
2. \(\lim\limits_{h\to 0}\tfrac{\sqrt[3]{8+h}-2}{h}\)

It will be helpful to recall that

\[ x^2-y^2=(x-y)(x+y)\quad\text{and}\quad x^3-y^3=(x-y)(x^2+xy+y^2). \]

1. Use the factorization for the difference of squares to notice that \((\sqrt{4+h}-2)(\sqrt{4+h}+2)=(\sqrt{4+h})^2-2^2=h.\) Rewrite the expression to evaluate the limit like this:

\[ \begin{align*}\lim_{h\to 0}\frac{\sqrt{4+h}-2}{h}&=\lim_{h\to 0}\frac{\sqrt{4+h}-2}{h}\cdot\frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}\\&=\lim_{h\to 0}\frac{h}{h(\sqrt{4+h}+2)}&&\text{expand numerator}\\&=\lim_{h\to 0}\frac{1}{\sqrt{4+h}+2}&&\text{reduce $\tfrac{h}{h}$ term}\\&=\frac{1}{\sqrt{4}+2}&&\text{evaluate limit}\\&=\frac{1}{4}.\end{align*} \]

2. Use the factorization for the difference of cubes to notice that \(\left(\sqrt[3]{8+h}-2\right)\left((8+h)^\frac{2}{3}+2(8+h)^\frac{1}{3}+2^2\right)=\left(\sqrt[3]{8+h}\right)^3-2^3=h.\) Rewrite the expression to evaluate the limit like this:

   \[ \begin{align*} \lim_{h\to 0}\frac{\sqrt[3]{8+h}-2}{h}&=\lim_{h\to 0}\frac{\sqrt[3]{8+h}-2}{h}\cdot\frac{(8+h)^\frac{2}{3}+2(8+h)^\frac{1}{3}+2^2}{(8+h)^\frac{2}{3}+2(8+h)^\frac{1}{3}+2^2}\\ &=\lim_{h\to 0}\frac{h}{h((8+h)^\frac{2}{3}+2(8+h)^\frac{1}{3}+2^2)}&&\text{expand numerator}\\ &=\lim_{h\to 0}\frac{1}{(8+h)^\frac{2}{3}+2(8+h)^\frac{1}{3}+2^2}&&\text{reduce $\tfrac{h}{h}$ term}\\ &=\frac{1}{8^\frac{2}{3}+2(8)^\frac{1}3+2^2}&&\text{evaluate limit}\\ &=\frac{1}{12}. \end{align*} \]

One Sided Limits

The function \(f\) that is given by \[f(x) = \tfrac{|x|}{x},\] looks like this:

Although \(\displaystyle\lim_{x\to 0}f(x)\) does not exist, much can still be said about the local properties of \(f\) as long as we restrict \(f\) to the left or to the right of \(0\).

This example motivates a refinement of the definition of a limit.

Limit from the Left and Limit from the Right

To say that the real number \(L\) is the limit of \(f(x)\) as \(x\) tends to \(x_0\) from the left means the following:

• For any sequence \((x_n)\) in \(\mathcal D(f)\cap (-\infty, x_0)\) that converges to \(x_0\), \((f(x_n))\) converges to \(L\). Express this symbolically as: \(\displaystyle \lim_{x\to x_0^-} f(x) = L\).

To say that the real number \(L\) is the limit of \(f(x)\) as \(x\) tends to \(x_0\) from the right means the following:

For any sequence \((x_n)\) in \(\mathcal D(f)\cap (x_0, \infty)\) that converges to \(x_0\), \((f(x_n))\) converges to \(L\). Express this symbolically as: \(\displaystyle \lim_{x\to x_0^+} f(x) = L\).

Understand the difference with the figure below.

The limits defined above are one sided limits.

All theorems about one sided limits (for example: limit laws, squeeze theorem) are also valid for one sided limits because one sided limits are limits for the function with domain restricted to the left or right of the limit point \(x_0\).

Limit Exist if and only if One Sided Limit Exist

Note: \(\displaystyle\lim_{x\to x_0} f(x)\) exists if and only if there is an \(L\) in \(\mathbb R\) so that \[\lim_{x\to x_0^-}f(x) = L\quad {\rm and}\quad \lim_{x\to x_0^+}f(x) = L.\] If it is true, then \(\displaystyle\lim_{x\to x_0} f(x)=L\)

Understand the theorem by using it in the next example.

Example 10

Calculate the following limits and carefully justify your reasoning:

1. \(\displaystyle \lim_{x\to 2^-} \tfrac{x^2+ 3x - 10}{x-2}\)

2. \(\displaystyle \lim_{x\to 5^-} \tfrac{|x-5|}{x-5}\)

3. \(\displaystyle \lim_{x\to 5^+} \tfrac{|x-5|}{x-5}\)

The limit laws do not apply because the limit of the denominator is zero. However, because the limit of the numerator is also zero, rewrite the expression so that the limit laws can apply.

1. Take the sequence \((x_n)\) in \((-\infty,2)\) that converges to \(2\). Rewrite \(\tfrac{(x_n)^2+ 3(x_n) - 10}{x_n-2}\) like this: \[\begin{align*} \tfrac{(x_n)^2+ 3(x_n) - 10}{x_n-2}&=\tfrac{(x_n-2)(x_n+5)}{x_n-2}\\ &=x_n+5, \end{align*}\] which holds for all \(n\) since \(x_n\not=2\). Therefore, \[\displaystyle \lim_{n\to \infty} \tfrac{(x_n)^2+ 3(x_n) - 10}{x-2}=\displaystyle \lim_{x\to \infty}(x_n+5)=2+5=7.\] Since this worked for any sequence in \((-\infty,2)\), the conclusion is that \[\displaystyle \lim_{x\to 2^-} \tfrac{x^2+ 3x - 10}{x-2}=7.\]

2. Take the sequence \((x_n)\) to be in \((-\infty,5)\) that converges to \(5\). Rewrite \(\tfrac{|x_n-5|}{x_n-5}\). Notice that \[|x_n-5|=\begin{cases} x_n-5&\text{ if }x_n-5\geq 0\\ -(x_n-5)&\text{ if }x_n-5< 0\\ \end{cases}\] or \[|x_n-5|=\begin{cases} x_n-5&\text{ if }x_n\geq 5\\ -(x_n-5)&\text{ if }x_n< 5\\ \end{cases}.\] Since \((x_n)\) in \((-\infty,5)\), this means \(x_n<5\) for all \(n\). Thus \[ \tfrac{|x_n-5|}{x_n-5}=\tfrac{-(x_n-5)}{x_n-5}=-1.\] which holds for all \(n\) since \(x_n\not=5\). Thus \[\displaystyle \lim_{n\to \infty}\tfrac{|x_n-5|}{x_n-5}=\displaystyle \lim_{n\to \infty}(-1)=-1.\] Since this holds for all sequences \((-\infty,5)\) that converge to \(5\), the conclusion is \[\displaystyle \lim_{x\to 5^-}\tfrac{|x-5|}{x-5}=-1.\]

3. Take the sequence \((x_n)\) to be in \((5,\infty)\) that converges to \(5\). Rewrite \(\tfrac{|x_n-5|}{x_n-5}\). Notice that \[|x_n-5|=\begin{cases} x_n-5&\text{ if }x_n-5\geq 0\\ -(x_n-5)&\text{ if }x_n-5< 0\\ \end{cases}\] or \[|x_n-5|=\begin{cases} x_n-5&\text{ if }x_n\geq 5\\ -(x_n-5)&\text{ if }x_n< 5\\ \end{cases}.\] Since \((x_n)\) in \((5,\infty)\), this means \(x_n>5\) for all \(n\). Thus \[ \tfrac{|x_n-5|}{x_n-5}=\tfrac{x_n-5}{x_n-5}=1,\] which holds for all \(n\) since \(x_n\not=5\). Thus \[\displaystyle \lim_{n\to \infty}\tfrac{|x_n-5|}{x_n-5}=\lim_{n\to \infty}1=1.\] Since this holds for all sequences \((5.\infty)\) that converge to \(5\), the conclusion is \[\displaystyle \lim_{x\to 5^+}\tfrac{|x-5|}{x-5}=1.\]

Now, let’s make sure we understand how to use graphical information to compute limits.

Example 11

Use the sketch of the function \(f\) to fill out the table below:

	\(\displaystyle\lim_{x\to a^-}f(x)\)	\(\displaystyle\lim_{x\to a^+}f(x)\)	\(\displaystyle\lim_{x\to a}f(x)\)	\(f(a)\)
\(a=-4\)
\(a=-2\)
\(a=1\)
\(a=5\)

	\(\displaystyle\lim_{x\to a^-}f(x)\)	\(\displaystyle\lim_{x\to a^+}f(x)\)	\(\displaystyle\lim_{x\to a}f(x)\)	\(f(a)\)
\(a=-4\)	\(-2\)	\(-2\)	\(-2\)	undefined
\(a=-2\)	\(-1\)	does not exist	does not exist	\(2\)
\(a=1\)	does not exist	\(-2\)	does not exist	\(-2\)
\(a=5\)	\(4\)	\(1\)	does not exist	\(1\)

Infinite Limits

Take \(f\) to be a real valued function on \(\mathbb R\) so that \(\mathcal D(f)\) is unbounded above.

The following statement about \(f\) makes precise the idea that, as long as \(x\) is large enough, \(f(x)\) is as close as we like to the real number \(L\):

Limit at Infinity

Take \(f\) to be a real valued function on \(\mathbb R\) so that \(\mathcal D(f)\) is unbounded above.

For any sequence \((x_n)\) in \(\mathcal D(f)\) that diverges to \(\infty\), \((f(x_n))\) converges to \(L\).

• The real number \(L\) is the limit of \(f(x)\) as \(x\) tends to \(\infty\).

• Express this symbolically as: \(\displaystyle \lim_{x\to \infty} f(x) = L\).

This limit is referred to as a limit at infinity.

Understand the definition with the figure below.

Determine the limit at infinity for the following examples.

Example 12

Evaluate the following limits:

\[\lim\limits_{x\to \infty} \tfrac{2x+3}{x+1}\]

Take \(x_n\) to be a sequence that diverges to \(\infty\). Note that \(\displaystyle\lim_{n\to\infty}x_n=\infty\) implies that \(\displaystyle\lim_{n\to\infty}\frac{1}{x_n}=0.\) Rewrite the expression to calculate the limits: \[ \begin{align*} \frac{2x_n+3}{x_n+1}&=\frac{2x_n+3}{x_n+1}\cdot\tfrac{\tfrac{1}{x_n}}{\tfrac{1}{x_n}}\\ &=\frac{2x_n\cdot \tfrac{1}{x_n}+3\cdot \tfrac{1}{x_n}}{x_n\cdot \tfrac{1}{x_n}+1\cdot \tfrac{1}{x_n}}\\ &=\frac{2+3\cdot \tfrac{1}{x_n}}{1+ \tfrac{1}{x_n}}.\\ \end{align*} \]

Therefore, \[ \begin{align*} \displaystyle\lim_{n\to\infty}\frac{2x_n+3}{x_n+1}&=\displaystyle\lim_{n\to\infty}\frac{2+3\cdot \tfrac{1}{x_n}}{1+ \tfrac{1}{x_n}}\\ &=\displaystyle\lim_{n\to\infty}\frac{2+3\cdot \tfrac{1}{x_n}}{1+ \tfrac{1}{x_n}}\\ &=\frac{\displaystyle\lim_{n\to\infty}2+\displaystyle\lim_{n\to\infty}3\cdot \tfrac{1}{x_n}}{\displaystyle\lim_{n\to\infty}1+ \displaystyle\lim_{n\to\infty}\tfrac{1}{x_n}}\\ &=\frac{2+3\cdot 0}{1+ 0}\\ &=\frac{2}{1}\\ &=2. \end{align*} \]

Since this holds for any sequence \(x_n\) that diverges to \(\infty\), the conclusion is \(\displaystyle \lim_{x\to \infty} \tfrac{2x+3}{x+1}=2.\)

Take \(f\) to be a real valued function on \(\mathbb R\) so that \(\mathcal D(f)\) is unbounded below.

The following statement about \(f\) makes precise the idea that, as long as \(-x\) is large enough, \(f(x)\) is as close as we like to the real number \(L\):

Limit at Negative Infinity

Take \(f\) to be a real valued function on \(\mathbb R\) so that \(\mathcal D(f)\) is unbounded below. For any sequence \((x_n)\) in \(\mathcal D(f)\) that diverges to \(-\infty\), \((f(x_n))\) converges to \(L\).

• The real number \(L\) is the limit of \(f(x)\) as \(x\) tends to \(\infty\).

• Express this symbolically as: \(\displaystyle \lim_{x\to -\infty} f(x) = L\)

This limit is referred to as a limit at negative infinity.

Use the figure below to understand the meaning of limit at negative infinity.

Calculate the limit at negative infinity of the following example.

Example 13

Calculate \(\displaystyle \lim_{x\to -\infty} \tfrac{\sqrt{1-x}}{x^3 +1}\).

Take \(f\) to be the function given by \(f(x)= \tfrac{\sqrt{1-x}}{x^3 +1}\) and take \((x_n)\) to be a sequence in \(D(f)=(-\infty,-1)\cup(-1,1]\) that diverges to \(-\infty\). So \(\displaystyle \lim_{n\to\infty}x_n=-\infty\) implies that \(\displaystyle\lim_{n\to\infty}\frac{1}{x_n}=0\). The limit laws cannot be used because the numerator and denominator diverge. Instead rewrite the expression like this: \[ \begin{align*} \frac{\sqrt{1-x_n}}{(x_n)^3 +1}&=\frac{\sqrt{1-x_n}}{(x_n)^3 +1}\cdot\frac{\tfrac{1}{(x_n)^3}}{\tfrac{1}{(x_n)^3}}\\ &=\frac{\sqrt{1-x_n}\cdot \tfrac{1}{(x_n)^3}}{(x_n)^3\cdot \tfrac{1}{(x_n)^3} +1\cdot \tfrac{1}{(x_n)^3}}\\ &=\frac{\sqrt{1-x_n}\cdot \left(-\sqrt{\tfrac{1}{(x_n)^6}}\right)}{(x_n)^3\cdot \tfrac{1}{(x_n)^3} +1\cdot \tfrac{1}{(x_n)^3}}\\ &=\frac{-\sqrt{1\cdot\frac{1}{(x_n)^6}-\frac{x_n}{(x_n)^6}}}{1 + \tfrac{1}{(x_n)^3}}\\ &=\frac{-\sqrt{\frac{1}{(x_n)^6}-\frac{1}{(x_n)^5}}}{1 + \tfrac{1}{(x_n)^3}}. \end{align*} \] Hence, \[\begin{align*} \lim_{n\to\infty}\frac{\sqrt{1-x_n}}{(x_n)^3 +1}&=\lim_{n\to\infty}\frac{-\sqrt{\frac{1}{(x_n)^6}-\frac{1}{(x_n)^5}}}{1 + \tfrac{1}{(x_n)^3}}.\\ &=\frac{0}{1}\\ &=0. \end{align*} \] Since this holds for any sequence \(x_n\) that diverges to \(-\infty\), the conclusion is \(\displaystyle \lim_{x\to -\infty} \tfrac{\sqrt{1-x}}{x^3 +1}=0.\)

The limit of a function \(f\) at a point \(x_0\) can fail to exist (or diverge as opposed to converge) in important ways.

The following statement about \(f\) makes precise the idea that, as long as \(x\) is close enough to \(x_0\) but not equal to \(x_0\), \(f(x)\) (or \(-f(x)\)) is as large as we like:

Diverges to Infinity or Negative Infinity at a Real Number

For any sequence \((x_n)\) in \(\mathcal D(f)\smallsetminus\{x_0\}\) that converges to \(x_0\), \((f(x_n))\) diverges to \(\pm\infty\).

• Express this symbolically as: \(\displaystyle \lim_{x\to x_0} f(x) = \pm\infty\).

Note: The \(\pm\) sign is to separately consider the case that \(f(x)\) is as large as we like or \(-f(x)\) is as large as we like. We use this notation to simplify the exposition.

Determine how the limits diverge with the next example.

Example 14

Determine the way in which the following limits diverge:

1. \(\displaystyle \lim_{x\to 0} \tfrac{1}{x^2}\)

2. \(\displaystyle \lim_{x\to 2} \tfrac{x-3}{|x-2|}\).

1. Take \(x_n\) to be a null sequence for which \(x_n\not=0\). Then \((x_n)^2\) is positive null sequence, so \(\lim\limits_{n\to\infty}\tfrac{1}{(x_n)^2}=\infty\). Since this is true for any null sequence for which \(x_n\not=0\), \[\displaystyle \lim_{x\to 0}\frac{1}{x^2}=\infty.\]

2. Take \(x_n\) to be a sequence for which \(\displaystyle\lim_{n\to\infty}x_n=2\) and \(x_n\not=2.\) Notice that \[|x-2|=\begin{cases}x_n-2&\text{ if }x_n> 2\\-(x_n-2)&\text{ if }x_n< 2\\\end{cases}\] so \[\frac{x_n-3}{|x_n-2|}=\begin{cases}\frac{x_n-3}{x_n-2}&\text{ if }x_n> 2\\-\frac{x_n-3}{x_n-2}&\text{ if }x_n< 2\\\end{cases}.\] If \(x_n<2\), then \(-\tfrac{x_n-3}{x_n-2}<0.\) If \(x_n>2\) and \(x_n<3\), then \(\tfrac{x_n-3}{x_n-2}<0\) . Therefore, as \(x_n\) tends to \(2\), \(\lim\limits_{n\to\infty}\tfrac{x_n-3}{|x_n-2|}=-\infty\). Since this is true for any sequence for which \(\lim\limits_{n\to\infty}x_n=2\) and \(x_n\not=2\), the conclusion is that \[ \displaystyle \lim_{x\to 2} \tfrac{x-3}{|x-2|}=-\infty. \]

The divergence to infinity or negative infinity can also happen with one sided limits.

Diverges to Infinity or Negative Infinity to the Left and to the Right

Write \(\displaystyle \lim_{x\to x_0^-} f(x) = \pm\infty\) to mean:

• For any sequence \((x_n)\) in \(\mathcal D(f)\cap(-\infty, x_0)\) that converges to \(x_0\), \((f(x_n))\) diverges to \(\pm\infty\).

Write \(\displaystyle \lim_{x\to x_0^+} f(x) = \pm\infty\) to mean:

• For any sequence \((x_n)\) in \(\mathcal D(f)\cap(x_0, \infty)\) that converges to \(x_0\), \((f(x_n))\) diverges to \(\pm\infty\).

Understand the meaning of this with definition visually with this figure.

Practice with this example.

Example 15

Determine the way in which the following limits diverge:

1. \(\displaystyle \lim_{x\to 1^+} \tfrac{1}{x-1}\)

2. \(\displaystyle \lim_{x\to 1^-} \tfrac{1}{x-1}\)

3. \(\displaystyle \lim_{x\to 1} \tfrac{1}{x-1}\)

4. \(\displaystyle \lim_{x\to 0} \tfrac{1}{x}\)

1. Take \(x_n\) to be a sequence in \((1,\infty)\) for which \(\displaystyle\lim_{n\to\infty}x_n=1\) and \(x_n\not=1.\) Since \(x_n>1\), this means \(x_n-1>0\). And since \(\displaystyle\lim_{n\to\infty}x_n=1\) implies that \(x_n-1\) is a null sequence, we have that \(\displaystyle\lim_{n\to\infty}\frac{1}{x_n-1}=\infty\). Thus \(\displaystyle\lim_{x\to 1^+}\frac{1}{x-1}=\infty.\)

2. Take \(x_n\) to be a sequence in \((-\infty,1)\) for which \(\displaystyle\lim_{n\to\infty}x_n=1\) and \(x_n\not=1.\) Since \(x_n<1\), this means \(x_n-1<0\). And since \(\displaystyle\lim_{n\to\infty}x_n=1\) implies that \(x_n-1\) is a null sequence, we have that \(\displaystyle\lim_{n\to\infty}\frac{1}{x_n-1}=-\infty\). Thus \(\displaystyle\lim_{x\to 1^-}\frac{1}{x-1}=-\infty.\)

3. Since \(\displaystyle\lim_{x\to 1^+}\frac{1}{x-1}=\infty\) and \(\displaystyle\lim_{x\to 1^-}\frac{1}{x-1}=-\infty\) this means the limit at \(x=1\) does not exist.

4. Follow a similar argument in a,b, and c, to conclude \(\displaystyle\lim_{x\to 0^+}\frac{1}{x}=\infty\) and \(\displaystyle\lim_{x\to 0^-}\frac{1}{x}=-\infty\), which means that the limit \(x=0\) does not exist.

Limits of functions at \(-\infty\) or \(\infty\) may also diverge to \(-\infty\) or \(\infty\).

Diverges to Infinity or Negative Infinity at Infinity

For any function \(f\) whose domain is unbounded above, write \(\displaystyle \lim_{x\to \infty} f(x) = \pm\infty\) to mean:

• For any sequence \((x_n)\) in \(\mathcal D(f)\) that diverges to \(\infty\), \((f(x_n))\) diverges to \(\pm\infty\).

For any function \(f\) whose domain is unbounded below, write \(\displaystyle \lim_{x\to -\infty} f(x) = \pm\infty\) to mean:

• For any sequence \((x_n)\) in \(\mathcal D(f)\) that diverges to \(-\infty\), \((f(x_n))\) diverges to \(\pm\infty\).

Here are two simple examples.

Example 16

We have the following:

1. \(\displaystyle \lim_{x\to \infty} x^3=\infty\)

2. \(\displaystyle \lim_{x\to -\infty} x^3=-\infty\)

The limit laws and squeeze theorem are also valid for convergent limits at infinity, and the proofs are essentially unchanged. These computational tools are very useful for determining limits at infinity and negative infinity. To determine that a limit is an infinite limit typically requires making an estimate.

Example 17

Compute the following limits and carefully justify your reasoning:

1. \(\displaystyle \lim_{x\to \infty} \tfrac{x^3-x+1}{x+2}\)

2. \(\displaystyle \lim_{x\to -\infty} \tfrac{x^2+7}{5x-1}\)

3. \(\displaystyle \lim_{x\to \infty} \tfrac{1-x^2}{x-3}\)

4. \(\displaystyle \lim_{x\to -\infty} \tfrac{5x^3+2x-1}{x^3-4x+7}\)

5. \(\displaystyle \lim_{x\to \infty} \tfrac{1}{\sqrt{x+1}}\cos(3x-5)\)

1. Rewrite the expression like this: \[\begin{align*} \tfrac{x^3-x+1}{x+2}&= \tfrac{x^3-x+1}{x+2}\cdot \frac{\tfrac{1}{x}}{\tfrac{1}{x}}\\ &=\frac{x^2-1+\frac{1}{x}}{1+\frac{2}{x}}. \end{align*} \] Since \(x\) diverges to \(\infty\), the denominator approaches 1 while the numerator diverges to \(\infty.\) Thus \[\displaystyle \lim_{x\to \infty} \tfrac{x^3-x+1}{x+2}=\infty.\]

2. Notice that \[\begin{align*} \tfrac{x^2+7}{5x-1}&= \tfrac{x^2+7}{5x-1}\cdot \frac{\tfrac{1}{x}}{\tfrac{1}{x}}\\ &=\frac{x+\frac{7}{x}}{5-\frac{1}{x}}. \end{align*} \] Since \(x\) diverges to \(-\infty\), denominator approaches 5 while the numerator diverges to \(-\infty.\) Thus \[\displaystyle \lim_{x\to \infty}\tfrac{x^2+7}{5x-1}=-\infty.\]

3. Notice that \[\begin{align*} \tfrac{1-x^2}{x-3}&= \tfrac{1-x^2}{x-3}\cdot \frac{\tfrac{1}{x}}{\tfrac{1}{x}}\\ &=\frac{\frac{1}{x}-x}{1-\frac{3}{x}}. \end{align*} \] Since \(x\) tends to \(\infty\), the denominator approaches 1 while the numerator diverges to \(-\infty.\) Thus \[\displaystyle \lim_{x\to -\infty} \tfrac{1-x^2}{x-3}=-\infty.\]

4. Notice that \[\begin{align*} \tfrac{5x^3+2x-1}{x^3-4x+7}&= \tfrac{5x^3+2x-1}{x^3-4x+7}\cdot \frac{\tfrac{1}{x^3}}{\tfrac{1}{x^3}}\\ &=\frac{5+\frac{2}{x^2}-\frac{1}{x^3}}{1-\frac{4}{x^2}+\frac{7}{x^3}}. \end{align*}\] Since \(x\) tends to \(-\infty\), the numerator tends to \(5\) while the denominator tends to \(1\). Thus, \[\displaystyle \lim_{x\to -\infty} \tfrac{5x^3+2x-1}{x^3-4x+7}=5\]

5. Use the squeeze theorem. Since the cosine function is bounded between \(-1\) and \(1\), we have that \[-1\leq \cos(3x-5)\leq 1,\] As long as \(x> -1\), \(\sqrt{x+1}\) is defined and positive, so \[-1\leq \cos(3x-5)\leq 1\] implies that \[-\frac{1}{\sqrt{x+1}}\leq \frac{1}{\sqrt{x+1}}\cos(3x-5)\leq \frac{1}{\sqrt{x+1}}\] for \(x>-1\). Since \(\displaystyle\lim_{x\to\infty}-\frac{1}{\sqrt{x+1}}=0\) and \(\displaystyle\lim_{x\to\infty}\frac{1}{\sqrt{x+1}}=0\), use the squeeze theorem to conclude \(\displaystyle \lim_{x\to \infty} \tfrac{1}{\sqrt{x+1}}\cos(3x-5)=0.\)

In the next example pay attention to how we rewrite the term to obtain the limit.

Example 18

Compute \(\displaystyle \lim_{x\to \infty} \big(\sqrt{x^2 + 3x+1} - x\big)\).

Rewrite the expression like this: \[ \begin{align*} \left(\sqrt{x^2+3x+1}-x\right)&=\left(\sqrt{x^2+3x+1}-x\right)\cdot \frac{\sqrt{x^2+3x+1}+x}{\sqrt{x^2+3x+1}+x}\\ &=\frac{x^2+3x+1-x^2}{\sqrt{x^2+3x+1}+x}\\ &=\frac{3x+1}{\sqrt{x^2+3x+1}+x}\\ &=\frac{3x+1}{\sqrt{x^2+3x+1}+x}\cdot \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{\sqrt{x^2}}}\\ &=\frac{3+\frac{1}{\sqrt{x^2}}}{\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}+1}.\\ \end{align*} \] Thus

\[ \begin{align*} \lim_{x\to\infty}(\sqrt{x^2+3x+1}-x)&=\lim_{x\to\infty}\frac{3+\frac{1}{\sqrt{x^2}}}{\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}+1}\\ &=\frac{3+0}{\sqrt{1+0+0}+1}\\ &=\frac{3}{2}. \end{align*} \]

The ideas from Chapter 3 and Chapter 4, in regards to horizontal and vertical asymptotes can be made precise using the limits.

Here is the definition of a horizontal asymptote.

Horizontal Asymptote

The horizontal line that passes through \((0, L)\) is a horizontal asymptote for a function \(f\) if either \[\lim_{x\to -\infty} f(x) = L\quad {\rm or}\quad \lim_{x\to \infty}f(x) = L.\] Although rational functions may have only one horizontal asymptote, in general, functions may have up to two horizontal asymptotes

Understand a horizontal asymptote with the following figure.

This is the definition of a vertical asymptote.

Vertical Asymptote

The vertical line that passes through \((x_0, 0)\) is a vertical asymptote for a function \(f\) if either \[\lim_{x\to x_0^-} f(x) = \pm\infty\quad {\rm or}\quad \lim_{x\to x_0^+}f(x) = \pm\infty.\] Although rational functions cannot be defined at a vertical asymptote, this is not true for functions in general.

Understand this definition with the figure below

Find vertical and horizontal asymptotes with the following example.

Example 19

Determine all horizontal and vertical asymptotes of the function \(f\) that is given by \[f(x) = \begin{cases} \frac{2x-1}{x+2}\big(1+\frac{1}{x}\sin(x)\big) &\text{if } x<-1 \;\text{and }x\ne -2\\[.1in]\frac{1}{x-1} &\text{if } -1\leq x < 1\\[.1in]\frac{1-5x}{x+3} &\text{if } x\ge 1.\end{cases}\]

Horizontal asymptotes can be found by calculating the limit as \(x\) diverges to \(\infty\) and limit as \(x\) diverges to \(-\infty\).

When \(x\geq 1\), \(f(x)=\tfrac{1-5x}{x+3}\). So

\[\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{1-5x}{x+3}=-5.\]

When \(x<-1\) and \(x\not=-2\), \(f(x)=\tfrac{2x-1}{x+2}\left(1+\tfrac{1}{x}\sin(x)\right)\). So \[\begin{align*}\lim_{x\to-\infty}\frac{2x-1}{x+2}\big(1+\frac{1}{x}\sin(x)\big) &=\lim_{x\to-\infty}\frac{2x-1}{x+2} \lim_{x\to-\infty}\left(1+\frac{\sin(x)}{x}\right)\\ &=2\cdot(1+0)\\ &=2. \end{align*}\]

This means that the lines \(y=-5\) and \(y=2\) are horizontal asymptotes of the function \(f\).

The vertical asymptotes of the function \(f\) are at \(x_0\) for which the function diverges to \(\infty\) or \(-\infty\).

When \(x<-1\) and \(x\not=-2\), \(f(x)=\tfrac{2x-1}{x+2}\big(1+\tfrac{1}{x}\sin(x)\big)\). The term \(\left(1+\tfrac{1}{x}\sin(x)\right)\) converges to a positive number as \(x\to -2\), but \(\tfrac{2x-1}{x+2}\) diverges to \(\infty\) if \(x\to -2^-\) and diverges to \(-\infty\) if \(x\to-2^+\). To summarize, \(\lim\limits_{x\to {-2}^+}f(x)=-\infty\) and \(\lim\limits_{x\to {-2}^-}f(x)=\infty.\) Therefore, \(x=2\) is a vertical asymptote.

When \(-1\leq x<1\), \(f(x)=\tfrac{1}{x-1}\). So \(\lim\limits_{x\to 1^+}\tfrac{1}{x-1}=\infty\) and \(\lim\limits_{x\to 1^-}\tfrac{1}{x-1}=-\infty.\) Therefore, \(x=1\) is a vertical asymptote.

To summarize,

• \(y=-5\) and \(y=2\) are horizontal asymptotes
• \(x=-2\) and \(x=1\) are vertical asymptotes.

Asymptotic behavior from Chapter 3 and 4 can also be restated using limits as well.

Asymptotic Behavior

Functions \(f\) and \(g\) have the same asymptotic behavior at \(\infty\) or \(-\infty\), respectively, if there is a constant \(L\) so that \[\lim_{x\to -\infty}\frac{f(x)}{g(x)} = L\quad {\rm or}\quad \lim_{x\to \infty}\frac{f(x)}{g(x)} = L.\]

When will a rational function have the same asymptotic behavior as \(C{\rm pow}_n\) for some nonzero real number \(C\) and some natural number \(n\)? See that with the next example.

Example 20

Take \(f\) to be the rational function that is given by \[f(x) = \tfrac{(x+5)^2(2x-3)^2(4-x)^3}{(3x-1)(x+8)^5}.\] Find a monomial \(g\) so that \(f\) and \(g\) have the same asymptotic behavior at both \(\infty\) and \(-\infty\)

The leading term of the numerator is \(x^2\cdot (2x)^2\cdot (-x)^3=-4x^7\). The leading term of the denominator is \(3x\cdot x^5=3x^6.\) The claim is that \(g(x)=\tfrac{-4x^7}{3x^6}=-\tfrac{4}{3}x\) will have the same asymptotic behavior at both \(\infty\) and \(-\infty\).

Notice that \[\frac{f(x)}{g(x)}=\frac{\tfrac{(x+5)^2(2x-3)^2(4-x)^3}{(3x-1)(x+8)^5}}{-\frac{4}{3}x}=\frac{\tfrac{-4x^7+\text{lower order terms}}{3x^6+\text{lower order terms}}}{-\frac{4}{3}x}=\tfrac{-4x^7+\text{lower order terms}}{-4x^7+\text{lower order terms}}.\] Multiply numerator and denominator by \(\tfrac{1}{x^7}\), the numerator and denominator will be \(-4\) and a sum of terms which tend to \(0\) as \(x\) diverges to \(\infty\) or \(-\infty\). Therefore \[\lim_{x\to\infty}\frac{f(x)}{g(x)}=\frac{-4}{-4}=1,\] which shows that\(f(x)=\tfrac{(x+5)^2(2x-3)^2(4-x)^3}{(3x-1)(x+8)^5}\) and \(g(x)=-\tfrac{4}{3}x\) both have the same asymptotic behavior at both \(\infty\) and \(-\infty\).

Knowing the asymptotic properties of an invertible function will determine the asymptotic properties of the function’s inverse.

It is helpful to first recognize the following facts:

Special Exponential and Tangent limits

\[\lim_{x\to \infty} \mathrm{e}^x = \infty, \; \lim_{x\to -\infty} \mathrm{e}^{x} = 0,\; \lim_{x\to \frac{\pi}{2}^-} \tan(x) = \infty,\; \text{and}\; \lim_{x\to -\frac{\pi}{2}^+} \tan(x) = -\infty.\]

Use these facts to understand how to compute the following limits.

Example 21

Determine the following limits:

1. \(\displaystyle \lim_{x\to \infty} \ln(x)\)

2. \(\displaystyle \lim_{x\to 0^+} \ln(x)\)

3. \(\displaystyle \lim_{x\to \infty} \arctan(x)\)

4. \(\displaystyle \lim_{x\to -\infty} \arctan(x)\).

Recall that for an invertible function \(f\), \(f(x)=y\) if and only if \(x=f(y)\), where \(x\in\mathcal{D}(f).\)

1. \(\displaystyle \lim_{x\to \infty} \ln(x)=\infty\);

2. \(\displaystyle \lim_{x\to 0^+} \ln(x)=-\infty\);

3. \(\displaystyle \lim_{x\to \infty} \arctan(x)=\frac{\pi}{2}\)

4. \(\displaystyle \lim_{x\to -\infty} \arctan(x)=-\frac{\pi}{2}\)

Limits and Paths

Everything we learned about taking limits of functions extends to the idea of taking limits of paths. Here is the definition.

Limit of a Path

For any \((x_0,y_0)\) in \(\mathbb{R}^2\) and any path \(c\) with component functions \(x\) and \(y\), that is, \[c(t)=(x(t),y(t)),\] these conditions are equivalent:

• (Condition 1) \(\displaystyle\lim_{t\to t_0}c(t)=(x_0,y_0)\)
• (Condition 2) \(\displaystyle\lim_{t\to t_0}x(t)=x_0\) and \(\displaystyle\lim_{t\to t_0}y(t)=y_0\)

Practice calculating limits of paths with the following example. When appropriate remember our limit laws and techniques for computing limits.

Example 22

Take \(c\) to be the path that is given for each \(t\) in \(\mathbb{R}\smallsetminus \{2\}\) by \[c(t)=\left(\frac{t}{t^2+1}\sin(t^2),\frac{t^2-4}{t-2}\right).\]

Determine the following limits:

1. \(\displaystyle\lim_{t\to 2^-}c(t)\)
2. \(\displaystyle\lim_{t\to 2^{+}}c(t)\)
3. \(\displaystyle\lim_{t\to -\infty}c(t)\)
4. \(\displaystyle\lim_{t\to \infty}c(t)\)

In this problem, the component functions are \(x(t)=\frac{t}{t^2+1}\sin(t^2)\) and \(y(t)=\frac{t^2-4}{t-2}.\)

1. Note that \[\begin{align*}\lim_{t\to 2^-}x(t)&=\lim_{t\to 2^-}\frac{t}{t^2+1}\sin(t^2)\\ &=\frac{2}{2^2+1}\sin(2^2)\\ &=\frac{2}{5}\sin(4) \end{align*}\] and \[\lim_{t\to 2^{+}}y(t)=\lim_{t\to 2^{+}}\frac{t^2-4}{t-2}=4.\] Therefore, \[\lim_{t\to 2^-}c(t)=\left(\frac{2}{5}\sin(4),4\right).\]

2. Use a similar argument from above to obtain that \[\lim_{t\to 2^+}c(t)=\left(\frac{2}{5}\sin(4),4\right).\]

3. Use the squeeze theorem to obtain that \[\lim_{t\to -\infty}x(t)=\lim_{t\to -\infty}\frac{t}{t^2+1}\sin(t^2)=0.\] Also notice that \[\lim_{t\to -\infty}y(t)=\lim_{t\to -\infty}\frac{t^2-4}{t-2}=-\infty.\] Therefore, \[\lim_{t\to -\infty}c(t)=\left(0,-\infty\right).\]

4. Use the squeeze theorem to obtain that \[\lim_{t\to \infty}x(t)=\lim_{t\to \infty}\frac{t}{t^2+1}\sin(t^2)=0.\] Also notice that \[\lim_{t\to \infty}y(t)=\lim_{t\to \infty}\frac{t^2-4}{t-2}=\infty.\] Therefore, \[\lim_{t\to \infty}c(t)=\left(0,\infty\right).\]

Using what we know about limits can help us construct paths with special properties.

Example 23

Identify a path that describes the position in time of a particle that moves along the line segment \(L\) with endpoints \((1,-5)\) and \((4,3)\), has domain equal to \(\mathbb{R}\), is at the midpoint of \(L\) at time \(0\), moves only to the right, and reaches all points of \(L\) except the endpoints of \(L\).

There are many different answers. Here is one example.

First the vector that moves points along this line: \[V=(4,3)-(1,-5)=\langle 3,8\rangle.\]

The midpoint of the path is

\[\frac{1}{2}V+(1,-5)=\left\langle \frac{3}{2},4\right\rangle+(1,-5)=\left(\frac{5}{2},-1\right).\]

The path of the particle \(c\) will have this form

\[c(t)=\frac{1}{2}d(t)V+\left(\frac{5}{2},-1\right)\]

where \(d\) is a function so that the following is true:

\[d(0)=0,\quad \lim_{t\to\infty}d(t)=1,\quad\text{and}\quad \lim_{t\to-\infty}d(t)=-1.\]

We know that \[\lim_{t\to\infty}\arctan(t)=\frac{\pi}{2}\quad\text{and}\quad\lim_{t\to-\infty}\arctan(t)=-\frac{\pi}{2}\] so take \[d(t)=\frac{2}{\pi}\arctan(t)\] so that \[\lim_{t\to\infty}d(t)=1\quad\text{and}\quad\lim_{t\to-\infty}d(t)=-1.\]

The path is thus

\[c(t)=\frac{1}{\pi}\arctan(t)\langle 3,8\rangle+\left(\frac{5}{2},-1\right).\]

Here is another example.

Example 24

Identify a path that describes the position in time of a particle that has domain equal to \((0,2),\) that moves along the line \(L\) with slope \(3\) that intersects \((4,1),\) that moves only to the right along \(L\), that is \((4,1)\) at time \(1\), and that reaches every point of \(L\). Simulate the motion of this particle.

Here is one example of such a path. Take \(c\) to be the path given by

\[ c(t)=d(t-1)\langle 1,3\rangle +(4,1), \]

where \(d\) is a function defined on \((-1,1)\) with \[\lim_{t\to-1}d(t)=-\infty\quad\text{and}\quad\lim_{t\to 1}d(t)=\infty.\]

Here is such an example of a function \(d\): \[d(t)=\tan\left(t\tfrac{\pi}{2}\right).\] Therefore, our path is \[c(t)=\tan\left((t-1)\frac{\pi}{2}\right)\langle 1,3\rangle +(4,1).\]

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